My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum

LinkBack Thread Tools Display Modes
June 16th, 2010, 05:57 PM   #1
Joined: Feb 2009

Posts: 76
Thanks: 0

The range of a matrix transformation

How to do you determine if a vector (w) is in the range a matrix transformation.

Let f: R^2 --> R^3
f(x) = Ax

Vector w

Why is it a no?
My work: ( I don't know if I'm doing it right)

and why is for Vector w, it's a yes?

The book says that the set of all images of the vectors in R^n is called the range of f. I don't know what it means. Can someone please explain to me how to determine the range. Thanks in advance
remeday86 is offline  
June 16th, 2010, 07:21 PM   #2
Senior Member
Joined: Oct 2007
From: Chicago

Posts: 1,701
Thanks: 3

Re: The range of a matrix transformation

You want w= f(v) = Av, where v is a vector .

The equation should be:

You cannot substitute values of w into v, you need to see if any value of v, when multiplied by A gives you w. See if there is any v=[x,y] which satisfies this equation.

Edit: Some clarification. "The set of all images" is the set . So, if w is in the image, w=f(v) for some vector v in R^2. To check if such a vector exists, see if there is any vector which satisfies f(v)=w.
cknapp is offline  

  My Math Forum > College Math Forum > Linear Algebra

matrix, range, transformation

Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Getting Transformation Matrix tomtong Algebra 0 June 10th, 2013 05:53 PM
Looking for the right transformation matrix Niko Bellic Linear Algebra 5 January 3rd, 2013 11:32 AM
Matrix Transformation aliya Linear Algebra 1 August 10th, 2010 12:39 PM
Kernal and range of a linear transformation remeday86 Linear Algebra 3 July 25th, 2010 04:44 PM
Range and Kernel of a Linear Transformation sansar Linear Algebra 1 April 7th, 2009 10:07 PM

Copyright © 2017 My Math Forum. All rights reserved.